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Boundary Value Problems
Volume 2007, Article ID 48348, 20 pages
doi:10.1155/2007/48348
Research Article
Unbounded Supersolutions of Nonlinear Equations with
Nonstandard Growth
Petteri Harjulehto, Juha Kinnunen, and Teemu Lukkari
Received 3 March 2006; Revised 16 May 2006; Accepted 28 May 2006
Recommended by Ugo Pietro Gianazza
We show that every weak supersolution of a variable exponent p-Laplace equation is
lower semicontinuous and that the singular set of such a function is of zero capacity if the
exponent is logarithmically H
¨
older continuous. As a technical tool we derive Harnack-
type estimates for possibly unbounded supersolutions.
Copyright © 2007 Petteri Harjulehto et al. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1. Introduction
The purpose of this work is to study regularity theory related to partial di fferential equa-
tions with nonstandard growth conditions. The principal prototype that we have in mind
is the equation
div

p(x)


∇
u(x)



p(x)−2
∇u(x)

=
0, (1.1)
which is the Euler-Lagrange equation of the variational integral



∇
u(x)


p(x)
dx. (1.2)
Here p(
·) is a measurable function satisfying
1 < inf
x∈R
n
p(x) ≤ p(x) ≤ sup
x∈R
n
p(x) < ∞. (1.3)
If p(
·) is a constant function, then we have the standard p-Laplace equation and p-
Dirichlet integral. This kind of variable exponent p-Laplace equation has first been con-
sidered by Zhikov [1] in connection with the Lavrentiev phenomenon for a Thermistor
2 Boundary Value Problems

problem. By now there is an extensive literature on partial differential equations with
nonstandard growth conditions; for example, s ee [2–6].
It has turned out that regularity results for weak solutions of (1.1) do not hold without
additional assumptions on the variable exponent. In [1] Zhikov introduced a logarith-
mic condition on modulus of continuity. Variants of this condition have been expedient
tools in the study of maximal functions, singular integral operators, and partial differen-
tial equations with nonstandard growth conditions on variable exponent spaces. Under
this assumption Harnack’s inequality and local H
¨
older continuity follow from Moser or
DeGiorgi-type procedure; see [7, 8].Seealso[9]. An interesting feature of this theory is
that estimates are intrinsic in the sense that they depend on the solution itself. For exam-
ple, supersolutions are assumed to be locally bounded and Harnack-type estimates in [7]
depend on this bound.
In this work we are interested in possibly unbounded supersolutions of (1.1) and hence
the previously obtained estimates are not immediately available for us. The main nov-
elty of our approach is that instead of the boundedness we apply summability estimates
for supersolutions. Roughly speaking we are able to replace L
∞
-estimates with certain
L
p
-estimates for small values of p. The argument is a modification of Moser’s iteration
scheme presented in [7]. However, the modification is not completely straightforward
and we have chosen to present all details here. As a by-product, we obtain refinements of
results in [7, 9].
After these technical adjustments we are ready for our main results. Solutions are
known to b e continuous and hence it is natural to ask whether supersolutions are semi-
continuous. Indeed, using Harnack-type estimates we show that every supersolution has
a lower semicontinuous representative. Thus it is possible to study pointwise behavior of

supersolutions. Our main result states that the singular set of a supersolution is of zero
capacity. For the capacity theory in variable exponent spaces we refer to [10]. In fact we
study a slightly more general class of functions than supersolutions which corresponds to
the class of superharmonic functions in the case when p(
·)isconstant;see[11, 12].
2. Preliminaries
A measurable function p :
R
n
→ (1,∞) is called a variable exponent. We denote
p
+
A
= sup
x∈A
p(x), p
−
A
= inf
x∈A
p(x), p
+
= sup
x∈R
n
p(x), p
−
= inf
x∈R
n

p(x) (2.1)
and assume that 1 <p
−
≤ p
+
< ∞.
Let Ω be an open subset of
R
n
with n ≥ 2. The variable exponent Lebesgue space
L
p(·)
(Ω) consists of all measurable functions u defined on Ω for which

Ω


u(x)


p(x)
dx<∞. (2.2)
The Luxemburg norm on this space is defined as
u
p(·)
= inf

λ>0:

Ω





u(x)
λ




p(x)
dx ≤ 1

. (2.3)
Petteri Harjulehto et al. 3
Equipped with this norm L
p(·)
(Ω) is a Banach space. T he variable exponent Lebesgue
space is a special case of a more general Orlicz-Musielak space studied in [13]. For a
constant function p(
·) the variable exponent Lebesgue space coincides with the standard
Lebesgue space.
The var iable exponent Sobolev space W
1,p(·)
(Ω) consists of functions u ∈ L
p(·)
(Ω)
whose distributional gradient
∇u exists almost everywhere and belongs to L
p(·)

(Ω). The
variable exponent Sobolev space W
1,p(·)
(Ω) is a Banach space with the norm
u
1,p(·)
=u
p(·)
+ ∇u
p(·)
. (2.4)
For basic results on variable exponent spaces we refer to [14]. See also [15].
A somewhat unexpected feature of the variable exponent Sobolev spaces is that smooth
functions need not be dense without additional assumptions on the variable exponent.
This was observed by Zhikov in connection with the so-called Lavrentiev phenomenon.
In [1] he introduced a logarithmic condition on modulus of continuity of the variable
exponent. Next we briefly recall a version of this condition. The variable exponent p is
said to satisfy a logarithmic H
¨
older continuity property, or briefly log-H
¨
older, if there is
aconstantC>0suchthat


p(x) −p(y)


≤
C

−log

|
x − y|

(2.5)
for all x, y
∈ Ω such that |x −y|≤1/2. Under this condition smooth functions are dense
in variable exponent Sobolev spaces and there is no confusion to define the Sobolev space
with zero boundary values W
1,p(·)
0
(Ω) as the completion of C
∞
0
(Ω) with respect to the
norm
u
1,p(·)
.Wereferto[16, 17]forthedetails.
In this work we do not need any deep properties of variable exponent spaces. For
our purposes, one of the most important facts about the variable exponent Lebesgue
spaces is the following. If E is a measurable set with a finite measure, and p and q are
variable exponents satisfying q(x)
≤ p(x) for almost every x ∈ E,thenL
p(·)
(E)embeds
continuously into L
q(·)
(E). In particular this implies that every function u ∈ W

1,p(·)
(Ω)
also belongs to W
1,p
−
Ω
loc
(Ω)andtoW
1,p
−
B
(B), where B ⊂ Ω is a ball. For all these facts we
refer to [15, 14].
We say that a function u
∈ W
1,p(·)
loc
(Ω) is a weak solution (supersolution) of (1.1), if

Ω
p(x)


∇
u


p(x)−2
∇u ·∇ϕdx = (≥)0 (2.6)
for e very test function ϕ

∈ C
∞
0
(Ω)(ϕ ≥ 0). When 1 <p
−
≤ p
+
< ∞ the dual of L
p(·)
(Ω)
is the space L
p

(·)
(Ω) obtained by conjugating the exponent pointwise, see [14]. This to-
gether with our definition W
1,p(·)
0
(Ω)asthecompletionofC
∞
0
(Ω) implies that we can
also test with functions ϕ
∈ W
1,p(·)
0
(Ω).
Our notation is r ather standard. Various constants are denoted by C and the value
of the constant may differ even on the same line. The quantities on which the constants
depend are given in the statements of the theorems and lemmas. A dependence on p

4 Boundary Value Problems
includes dependence on the log-H
¨
older-constant of p.Notealsothatduetothelocal
nature of the estimates, the constants depend only on the values of p in some ball.
3. Harnack estimates
In this section we prove a weak Harnack inequality for supersolutions. Throughout this
section we write
v
α
= u + R
α
, (3.1)
where u is a nonnegative supersolution.
We derive a suitable Caccioppoli-type estimate with variable exponents. Our aim is
to combine this estimate with the standard Sobole v inequality. Thus we need a suitable
passage between constant and variable exponents. This is accomplished in the following
lemma.
Lemma 3.1. Let E be a measurable subset of
R
n
. For all nonnegative measurable functions
f and g de fined on E,

E
fg
p
−
E
dx ≤


E
f dx +

E
fg
p(x)
dx. (3.2)
Proof. The claim follows from an integration of the pointwise inequality
f (x)g(x)
p
−
E
≤ f (x)+ f (x)g(x)
p(x)
. (3.3)
If p(x)
= p
−
E
this is immediate. Otherwise we apply Young’s inequality with the exponent
p(x)/p
−
E
> 1. 
Lemma 3.2 (Caccioppoli estimate). Suppose that u is a nonnegative supersolution in B
4R
.
Let E be a measurable subset of B
4R

and η ∈ C
∞
0
(B
4R
) such that 0 ≤ η ≤ 1. Then for every
γ
0
< 0 the re is a constant C depending on p and γ
0
such that the inequality

E
v
γ−1
α
|∇u|
p
−
E
η
p
+
B
4R
dx ≤ C

B
4R


η
p
+
B
4R
v
γ−1
α
+ v
γ+p(x)−1
α
|∇η|
p(x)

dx (3.4)
holds for every γ<γ
0
< 0 and α ∈R.
Proof. Let s
= p
+
B
4R
. We want to test with the function ψ = v
γ
α
η
s
. To this end we show
that ψ

∈ W
1,p(·)
0
(B
4R
). Since η has a compact support in B
4R
, it is enough to show that
ψ
∈ W
1,p(·)
(Ω). We observe that ψ ∈L
p(·)
(Ω) since |v
γ
α
|η
s
≤ R
αγ
.Furthermore,wehave
|∇ψ|≤


γv
γ−1
α
η
s
∇u + v

γ
α
sη
s−1
∇η


≤|
γ|R
α(γ−1)
|∇u|+ sR
αγ
|∇η|, (3.5)
from which we conclude that
|∇ψ|∈L
p(·)
(Ω).
Petteri Harjulehto et al. 5
Using the facts that u is a supersolution and ψ is a nonnegative test function we find
that
0
≤

B
4R
p(x)


∇
u(x)



p(x)−2
∇u(x) ·∇ψ(x)dx
=

B
4R
p(x)γ|∇u|
p(x)
η
s
v
γ−1
α
dx +

B
4R
p(x)s|∇u|
p(x)−2
v
γ
α
η
s−1
∇u ·∇η dx.
(3.6)
Since γ is a negative number, this implies



γ
0


p
−
B
4R

B
4R
|∇u|
p(x)
η
s
v
γ−1
α
dx ≤ s

B
4R
p(x)|∇u|
p(x)−2
v
γ
α
η
s−1

∇u ·∇η dx.
(3.7)
We denote the right-hand side of (3.7)byI. Since the left-hand side of (3.7)isnonnega-
tive, so is I. Using the ε-version of Young’s inequality we obtain
I
≤ s

B
4R
p(x)|∇u|
p(x)−1
v
γ
α
η
s−1
|∇η|dx
≤ s

B
4R

1
ε

p(x)−1
p(x)

v
(γ+p(x)−1)/p(x)

α
|∇η|η
s−s/p

(x)−1

p(x)
p(x)
+ εp(x)

|∇
u|
p(x)−1
η
s/p

(x)
v
γ−(γ+p(x)−1)/p(x)
α

p

(x)
p

(x)
dx
≤ s


1
ε

s−1

B
4R
v
γ+p(x)−1
α
|∇η|
p(x)
η
s−p(x)
dx
+ s(s −1)ε

B
4R
|∇u|
p(x)
η
s
v
γ−1
α
dx.
(3.8)
By combining this with (3.7) we arrive at



γ
0


p
−
B
4R

B
4R
|∇u|
p(x)
η
s
v
γ−1
α
dx
≤ s

1
ε

s−1

B
4R
v

γ+p(x)−1
α
|∇η|
p(x)
η
s−p(x)
dx + s(s −1)ε

B
4R
|∇u|
p(x)
η
s
v
γ−1
α
dx.
(3.9)
6 Boundary Value Problems
By choosing
ε
= min

1,


γ
0



p
−
B
4R
2s(s −1)

(3.10)
we can absorb the last term in (3.9) to the left-hand side and obtain

B
4R
|∇u|
p(x)
η
s
v
γ−1
α
dx ≤ s

2s(s −1)


γ
0


p
−

B
4R
+1

s−1
2
|γ
0
|p
−
B
4R

B
4R
v
γ+p(x)−1
α
|∇η|
p(x)
dx.
(3.11)
Taking f
= v
γ−1
α
η
s
and g =|∇u| in Lemma 3.1 and using inequality (3.11)wehavethe
desired estimate.


In the proof of the Caccioppoli estimate we did not use any other assumption on the
variable exponent p except that 1 <p
−
≤ p
+
< ∞. From now on we assume the logarith-
mic H
¨
older continuity. This is equivalent to the following estimate:


B


−(p
+
B
−p
−
B
)
≤ C, (3.12)
where B
Ω is any ball; see for example [18].
The next two lemmas will be used to handle the right-hand side of the Caccioppoli
estimate.
Lemma 3.3. If the exponent p(
·) is log-H
¨

older continuous,
r
−p(x)
≤ Cr
−p
−
E
(3.13)
provided x
∈ E ⊂ B
r
.
Proof. For r
≥ 1wehaver
−p(x)
≤ r
−p
−
E
. Suppose then that 0 <r<1. Since E ⊂ B
r
implies
osc
E
p ≤ osc
B
r
p,weobtain
r
−p(x)

≤ r
−p
+
E
≤ r
−(osc
E
p)
r
−p
−
E
≤ r
−(osc
B
r
p)
r
−p
−
E
≤ Cr
−p
−
E
, (3.14)
where we used logarithmic H
¨
older continuity in the last inequality.


In the following lemma the bar red integral sign denotes the integr al average.
Lemma 3.4. Let f be a positive measurable function and assume that the exponent p(
·) is
log-H
¨
older continuous. Then
−

B
r
f
p
+
B
r
−p
−
B
r
dx ≤ Cf 
p
+
B
r
−p
−
B
r
L
s

(B
r
)
(3.15)
for any s>p
+
B
r
− p
−
B
r
.
Petteri Harjulehto et al. 7
Proof. Let q
= p
+
B
r
− p
−
B
r
.H
¨
older’s inequality implies
−

B
r

f
p
+
B
r
−p
−
B
r
dx ≤
C
r
n


B
r
1dx

1−q/s


B
r
f
s
dx

q/s
≤

C
r
n
r
n(1−q/s)
f 
q
L
s
(B
r
)
≤ Cf 
q
L
s
(B
r
)
.
(3.16)
AgainweusedthelogarithmicH
¨
older continuity in the last inequality.

Later we apply Lemma 3.4 with f = u
q

. In this case the upper bound written in terms
of u is

C
u
q

(p
+
B
r
−p
−
B
r
)
L
q

s
(B
r
)
. (3.17)
Now we have everything ready for the iteration. We write
Φ( f ,q,B
r
) =

−

B
r

f
q
dx

1/q
(3.18)
for a nonnegative measurable function f .
Lemma 3.5. Assume that u is a nonnegative supersolution in B
4R
and let R ≤ ρ<r≤ 3R.
Then the inequality
Φ

v
1
,qβ,B
r

≤
C
1/|β|

1+|β|

p
+
B
4R
/|β|


r
r −ρ

p
+
B
4R
/|β|
Φ

v
1
,
βn
n −1
,B
ρ

(3.19)
holds for every β<0 and 1 <q<n/(n
− 1). The constant C depends on n, p,andthe
L
q

s
(B
4R
)-norm of u with s>p
+
B

4R
− p
−
B
4R
.
Proof. In Lemma 3.2 we take E
= B
4R
and γ = β − p
−
B
4R
+1.Thenγ<1 − p
−
B
4R
and thus

B
4R
v
β−p
−
B
4R
1
|∇u|
p
−

B
4R
η
p
+
B
4R
dx ≤ C

B
4R

η
p
+
4R
v
β−p
−
B
4R
1
+ v
β−p
−
B
4R
+p(x)
1
|∇η|

p(x)

dx.
(3.20)
Next we take a cutoff function η
∈ C
∞
0
(B
r
)with0≤η ≤ 1, η = 1inB
ρ
,and


∇
η


≤
Cr
R(r −ρ)
. (3.21)
By Lemma 3.3 we have


∇
η



−p(x)
≤ CR
−p(x)

r
r −ρ

p
+
B
4R
≤ CR
−p
−
B
4R

r
r −ρ

p
+
B
4R
. (3.22)
8 Boundary Value Problems
Using inequality (3.20) with this choice of η we have
−

B

r




∇

v
β/p
−
B
4R
1
η
p
+
B
4R
/p
−
B
4R





p
−
B

4R
dx
≤ C
−

B
r
|β|
p
−
B
4R
v
β−p
−
B
4R
1
|∇u|
p
−
B
4R
η
p
+
B
4R
dx + C
−


B
r
v
β
1
η
p
+
B
4R
−p
−
B
4R
|∇η|
p
−
B
4R
dx
≤ C|β|
p
−
B
4R
−

B
r


η
p
+
B
4R
v
β−p
−
B
4R
1
+ v
β−p
−
B
4R
+p(x)
1
|∇η|
p(x)

dx + C
−

B
r
v
β
1

η
p
+
B
4R
−p
−
B
4R
|∇η|
p
−
B
4R
dx
≤ C

1+|β|

p
+
B
4R

−

B
r
η
p

+
B
4R
v
β−p
−
B
4R
1
dx +
−

B
r
v
β−p
−
B
4R
+p(x)
1
|∇η|
p(x)
dx +
−

B
r
v
β

1
|∇η|
p
−
B
4R
dx

.
(3.23)
Now the goal is to estimate each integral in the parentheses by

−

B
r
v
qβ
1
dx

1/q
. (3.24)
The first integral can be estimated with H
¨
older’s inequality. Since v
−p
−
B
4R

1
≤ R
−p
−
B
4R
,wehave
−

B
r
η
p
+
B
4R
v
β−p
−
B
4R
1
dx ≤

−

B
r
v
q(β−p

−
B
4R
)
1
dx

1/q
≤ R
−p
−
B
4R

−

B
r
v
qβ
1
dx

1/q
. (3.25)
By (3.22), H
¨
older’s inequality, and Lemma 3.4 for the second integral we have
−


B
r
v
β−p
−
B
4R
+p(x)
1
|∇η|
p(x)
dx
≤ CR
−p
−
B
4R

r
r −ρ

p
+
B
4R
−

B
r
v

β−p
−
B
4R
+p(x)
1
dx
≤ CR
−p
−
B
4R

r
r −ρ

p
+
B
4R

−

B
r
v
q

(p(x)−p
−

B
4R
)
1
dx

1/q


−

B
r
v
qβ
1
dx

1/q
≤ CR
−p
−
B
4R

r
r −ρ

p
+

B
4R

1+


v
1


q

(p
+
B
4R
−p
−
B
4R
)
L
q

s
(B
4R
)

1/q



−

B
r
v
qβ
1
dx

1/q
.
(3.26)
Finally, for the third integral we have by H
¨
older’s inequalit y,
−

B
r
v
β
1


∇
η



p
−
B
4R
dx ≤ CR
−p
−
B
4R

r
r −ρ

p
−
B
4R

−

B
r
v
qβ
1
dx

1/q
. (3.27)
Petteri Harjulehto et al. 9

Now we have arrived at the inequality
−

B
r



∇

v
β/p
−
B
4R
1
η
p
+
B
4R
/p
−
B
4R




p

−
B
4R
dx
≤ C

1+


β



p
+
B
4R

1+v
1

q

(p
+
B
4R
−p
−
B

4R
)
L
q

s
(B
4R
)

1/q

R
−p
−
B
4R

r
r −ρ

p
+
B
4R

−

B
r

v
qβ
1
dx

1/q
.
(3.28)
By the Sobolev inequality

−

B
r
|u|
na/(n−1)
dx

(n−1)/na
≤ CR

−

B
r
|∇u|
a
dx

1/a

, (3.29)
where u
∈ W
1,a
0
(B
r
)anda = p
−
B
4R
,and(3.28)weobtain

−

B
ρ
v
βn/(n−1)
1
dx

(n−1)/n
≤

C
−

B
r


v
β/p
−
B
4R
1
η
p
+
B
4R
/p
−
B
4R

np
−
B
4R
/(n−1)
dx

(n−1)/n
≤ CR
p
−
B
4R

−

B
r



∇

v
β/p
−
B
4R
1
η
p
+
B
4R
/p
−
B
4R




p
−

B
4R
dx
≤ C

1+|β|

p
+
B
4R

r
r −ρ

p
+
B
4R

−

B
r
v
qβ
1
dx

1/q

.
(3.30)
The claim follows from this since β is a negative number.

The next lemma is the crucial passage from positive exponents to negative exponents
in the Moser iteration scheme.
Lemma 3.6. Assume that u is a nonnegative supersolution in B
4R
and s>p
+
B
4R
− p
−
B
4R
. Then
there ex ist constants q
0
> 0 and C depending on n, p,andL
s
(B
4R
)-norm of u such that
Φ

v
1
,q
0

,B
3R

≤
CΦ

v
1
,−q
0
,B
3R

. (3.31)
Proof. Choose a ball B
2r
⊂ B
4R
and a cutoff function η ∈C
∞
0
(B
2r
)suchthatη =1inB
r
and |∇η|≤C/r.TakingE = B
r
and γ = 1 − p
−
B

r
in Lemma 3.2 we have
−

B
r


∇
logv
1


p
−
B
r
dx ≤ C

−

B
2r
v
−p
−
B
r
1
+

−

B
2r
v
p(x)−p
−
B
r
1
r
−p(x)
dx

. (3.32)
10 Boundary Value Problems
Using Lemmas 3.3 and 3.4 and the estimate v
−p
−
B
r
1
≤ R
−p
−
B
r
≤ r
−p
−

B
r
we have
−

B
r


∇
logv
1


p
−
B
r
dx ≤ C

r
−p
−
B
r
+ r
−p
−
B
2r

−

B
2r
v
p(x)−p
−
B
r
1
dx

≤
C

r
−p
−
B
r
+ r
−p
−
B
2r

1+v
1

p

+
B
4R
−p
−
B
4R
L
s
(B
4R
)

.
(3.33)
Let f
= logv
1
. By the Poincar
´
e inequalit y and the above estimate we obtain
−

B
r


f − f
B
r



dx ≤

r
p
−
B
r
−

B
r


∇
f


dx

1/p
−
B
r
≤ C

1+r
p
−

B
r
−p
−
B
2r

1+v
1

p
+
B
4R
−p
−
B
4R
L
s
(B
4R
)

1/p
−
B
r
.
(3.34)

Note that p
−
B
r
≥ p
−
B
2r
since B
r
⊂ B
2r
, so that the right-hand side of (3.34) is bounded.
The rest of the proof is standard. Since (3.34) holds for all balls B
2r
⊂ B
4R
, by the John-
Nirenberg lemma there exist positive constants C
1
and C
2
depending on the right-hand
side of (3.34)suchthat
−

B
3R
e
C

1
|f −f
B
3R
|
dx ≤ C
2
. (3.35)
Using (3.35) we can conclude that
(
−

B
3R
e
C
1
f
dx

−

B
3R
e
−C
1
f
dx


=

−

B
3R
e
C
1
( f −f
B
3R
)
dx

−

B
3R
e
−C
1
( f −f
B
3R
)
dx

≤


−

B
3R
e
C
1
|f −f
B
3R
|
dx

2
≤ C
2
2
,
(3.36)
which implies that

−

B
3R
v
C
1
1
dx


1/C
1
=

−

B
3R
e
C
1
f
dx

1/C
1
≤ C
2/C
1
2

−

B
3R
e
−C
1
f

dx

−1/C
1
= C
2/C
1
2

−

B
3R
v
−C
1
1
dx

−1/C
1
,
(3.37)
so that we can take q
0
= C
1
. 
Note that the exponent q
0

in Lemma 3.6 also depends on the L
s
(B
4R
)-norm of u.
More precisely, the constant C
1
obtained from the John-Nirenberg lemma is a universal
PetteriHarjulehtoetal. 11
constant div ided by the right-hand side of (3.34). Thus we have
q
0
=
C
C

+ u
p
+
B
4R
−p
−
B
4R
L
s
(B
4R
)

. (3.38)
The following weak Harnack inequality is the main result of this section. It applies also
for unbounded supersolutions.
Theorem 3.7 (weak Harnack inequality). Assume that u is a nonnegative supersolution in
B
4R
, 1 <q<n/(n −1) and s>p
+
B
4R
− p
−
B
4R
. Then

−

B
2R
u
q
0
dx

1/q
0
≤ C

essinf

B
R
u(x)+R

, (3.39)
where q
0
is the exponent from Lemma 3.6 and C depends on n, p, q,andL
q

s
(B
4R
)-norm
of u.
Remark 3.8. (1) The main difference compared to Alkhutov’s result in [7, 9]isthatthe
constant and the exponent depend on the L
q

s
(B
4R
)-norm of u instead of the essential
supremum of u in B
4R
. This is a crucial advantage for us since we are interested in super-
solutions which may be unbounded.
(2) Since the exponent p(
·) is uniformly continuous, we can take for example q


s =
p
−
Ω
by choosing R small enough. Thus the constants in the estimates are finite for all
supersolutions u in a scale that depends only on p(
·).
Proof. Let R
≤ ρ<r≤ 3R, r
j
= ρ +2
−j
(r −ρ), and
ξ
j
=−

n
(n −1)q

j
q
0
(3.40)
for j
= 0,1,2, ByLemma 3.5 we hav e
Φ

v
1

,ξ
j
,B
r
j

≤
C
1/|ξ
j
|

1+


ξ
j



p
+
B
4R
/|ξ
j
|

r
j

r
j
−r
j+1

p
+
B
4R
/|ξ
j
|
Φ

v
1
,ξ
j+1
,B
r
j+1

. (3.41)
An iteration of this inequality yields
Φ

v
1
,−q
0

,B
r

≤
∞

j=0
C
1/|ξ
j
|

1+


ξ
j



p
+
B
4R
/|ξ
j
|

r
j

r
j
−r
j+1

p
+
B
4R
/|ξ
j
|
essinf
x∈B
ρ
v
1
(x)
≤ C

∞
j=0
1/|ξ
j
|
2

∞
j=0
jp

+
B
4R
/|ξ
j
|

r
r −ρ


∞
j=0
p
+
B
4R
/|ξ
j
|
×
∞

j=0

1+


ξ
j




p
+
B
4R
/|ξ
j
|
essinf
x∈B
ρ
v
1
(x) .
(3.42)
12 Boundary Value Problems
We estimate the remaining product by using the fact that
|ξ
j
| > 1whenj>j
0
and |ξ
j
|≤1
when j
≤ j
0
for some j

0
. This implies that
∞

j=0

1+


ξ
j



p
+
B
4R
/|ξ
j
|
≤ 2

j
0
j
=0
p
+
B

4R
/|ξ
j
|
2

∞
j=j
0
+1
p
+
B
4R
/|ξ
j
|

n
(n −1)q

p
+
B
4R
q
0

∞
j=j

0
+1
j((n−1)q/n)
j
≤ 2

∞
j=0
p
+
B
4R
/|ξ
j
|

n
(n −1)q

p
+
B
4R
q
0

∞
j=0
j((n−1)q/n)
j

.
(3.43)
All the series in the above estimates are convergent by the root test, so we obtain
Φ

v
1
,−q
0
,B
r

≤
C essinf
x∈B
ρ
v
1
(x) . (3.44)
Next we choose ρ
= R and r =3R and use Lemma 3.6 to get
Φ

v
1
,q
0
,B
3R


≤
C essinf
x∈B
R
v
1
(x) . (3.45)
Finally we observe that
Φ

v
1
,q
0
,B
2R

≤
CΦ

v
1
,q
0
,B
3R

. (3.46)
This completes the proof.


Lemma 3.4 can be u sed in the proof of the supremum estimate in [7] in the same way
as in the proof of Lemma 3.5. Combining this with the weak Harnack inequality above
one obtains the full Harnack inequality with the constant depending on the L
q

s
(B
4R
)-
norm of the solution instead of the supremum. This implies the local H
¨
older continuity
of solutions by the standard technique; see [19]. Summing up, we have the following
theorem.
Theorem 3.9 (the Harnack inequality). Let u be a nonnegative solution in B
4R
, 1 <q<
n/(n
−1),ands>p
+
B
4R
− p
−
B
4R
. Then
esssup
x∈B
R

u(x) ≤ C

essinf
x∈B
R
u(x)+R

, (3.47)
where the constant C depends on n, p,andtheL
q

s
(B
4R
)-norm of u.
The main difference compared to earlier results is that the constant depends on the
L
q

s
-norm instead of the essential supremum. The following example shows that the con-
stant in the Harnack inequality cannot be independent of u even if the exponent is Lips-
chitz continuous.
PetteriHarjulehtoetal. 13
Example 3.10. Let p: (0,1)
→ (1,∞)bedefinedby
p(x)
=
⎧
⎪

⎪
⎪
⎨
⎪
⎪
⎪
⎩
3for0<x≤
1
2
,
3
−2

x −
1
2

for
1
2
<x<1.
(3.48)
Suppose that u
a
∈ W
1,p(·)
(0,1) is the minimizer of the Dirichlet energy integral with the
boundary values 0 and a>0. Then u
a

is a solution with the same boundary values by [20,
Theorem 5.7].
Theorem 3.2 of [21]gives
u
a
(x) =

x
0

C
a
p(y)

1/(p(y)−1)
dy, (3.49)
where C
a
is a constant obtained from the equation

1
0

C
a
p(y)

1/(p(y)−1)
dy =a. (3.50)
Note that if a

→∞,thenC
a
→∞.In(0,1/2) the minimizer is linear, u
a
(x) =

(C
a
/3)x.In
(1/2,3/5) the gradient of u
a
increases from

C
a
/3to(5C
a
/14)
5/9
.In11/20, the midpoint
of (1/2,3/5), the gradient of u
a
is (10C
a
/29)
10/19
.Hencewefindthat
u
a


3
5

≥

C
a
3
1
2
+
1
20

10C
a
29

10/19
. (3.51)
Let B = B(1/2,1/10) = (2/5,3/5).Thenweobtain
esssup
x∈B


u
a
(x)



essinf
x∈B


u
a
(x)


≥

(C
a
/3)(1/2) + (1/20)(10C
a
/29)
10/19

(C
a
/3)(2/5)
=
5
4
+
1
8
1
√
3


10
29

10/19
C
1/38
a
−→ ∞
(3.52)
as a
→∞.
This example can be extended to the planar case by studying functions f
a
(x, y) = u
a
(x)
in
{(x, y):0<x<1, 0 <y<1} with the exponent q(x, y) = p(x).
4. The singular set of a supersolution
First we prove that every supersolution has a lower semicontinuous representative if the
exponent p(
·)islog-H
¨
older. For this purpose, we need the fact that supersolutions are
locally bounded from below. This is true because subsolutions are locally bounded above,
which can be seen from the proof of Theorem 1 in [7].
We set
u
∗

(x) = essliminf
y→x
u(y) = lim
r→0
essinf
y∈B(x,r)
u(y). (4.1)
14 Boundary Value Problems
Theorem 4.1. Let u be a function defined on Ω such that
(1) u is finite almost everywhere, and
(2) min
{u,λ} is a supersolution for every λ>0.
Then u
∗
is lower semicontinuous and
u
∗
(x) = u(x) for almost every x ∈ Ω. (4.2)
Remark 4.2. Obser ve that all supersolutions satisfy the assumptions of the previous the-
orem. We present the result in a slightly more general case, since we would like to include
functions which are increasing limits of supersolutions. For bounded supersolutions the
theorem has been studied in [9].
Proof. Let Ω

Ω and first assume that u is bounded above. Pick a point x ∈ Ω

,choose
R such that B(x,2R)
⊂ Ω


and let
M
= esssup
Ω

u +1. (4.3)
For 0 <r
≤ 2R denote m(r) = essinf
y∈B(x,r)
u(y). Since supersolutions are locally
bounded below, we have m(r) >
−∞ for 0 <r≤ 2R.
The function u
∗
is lower semicontinuous since u
∗
r
(x) = essinf
y∈B(x,r)
u(y)islower
semicontinuous and u
∗
is an increasing limit of the functions u
∗
r
.
We will complete the proof for bounded functions u by showing that
u
∗
(x) = lim

r→0
−

B(x,r)
u(y)dy. (4.4)
For every 0 < 5r
≤ R the function u −m(5r) is a nonnegative supersolution in B(x,4r).
Thus the weak Harnack inequality implies that
m(r)
−m(5r) ≥ C

−

B
2r

u −m(5r)

q
0
dx

1/q
0
−r

≥
C



M −m(5r)

(q
0
−1)/q
0

−

B
2r

u −m(5r)

dx

1/q
0
−r

,
(4.5)
where we assumed that q
0
< 1. This implies that
0
≤
−

B(x,2r)

udy −m(5r)
≤ C

M −m(5r)

1−q
0

m(r) −m(5r)+Cr

q
0
.
(4.6)
Since m(r)
−m(5r)+Cr tends to zero as r → 0, the above estimate implies (4.4).
For the general case, denote u
i
= min{u,i} for i = 1,2, and observe that
u
∗
(x) = lim
i→∞
u
∗
i
(x) . (4.7)
PetteriHarjulehtoetal. 15
To see that u
= u

∗
almost everywhere, consider the sets
E
=

x ∈ Ω : u(x) < ∞,u
∗
(x) = u(x)

,
F
=

x ∈ Ω : u(x) =∞, u
∗
(x) = u(x)

.
(4.8)
Then
|F|=0 since u is assumed to be finite almost everywhere. For the set E we hav e
E
⊂∪
i
E
i
,where
E
i
=


x ∈ Ω : u
∗
i
(x) = u
i
(x)

, (4.9)
|E
i
|=0 by the first part of the theorem, and the claim follows. 
Our next goal is to obtain estimates for t he singular set of a supersolution. To this end,
we derive two Caccioppoli-type estimates for a supersolution.
Lemma 4.3. Let u beanonnegativesupersolutioninB
4R
, η ∈ C
∞
0
(B
4R
) such that 0 ≤ η ≤ 1
and γ<γ
0
< 0. Then there is a constant C depending on p and γ
0
such that

B
4R



∇
u


p(x)
η
p
+
B
4R
u
γ−1
dx ≤ C

B
4R
u
γ+p(x)−1


∇
η


p(x)
dx. (4.10)
Proof. Denote u
k

= u +1/k. Testing w ith η
p
+
B
4R
u
γ
k
gives

B
4R


∇
u


p(x)
η
p
+
B
4R
u
γ−1
k
dx ≤ C

B

4R
u
γ+p(x)−1
k


∇
η


p(x)
dx (4.11)
as in the proof of inequality (3.11)inLemma 3.2. u
γ−1
k
→ u
γ−1
monotonically as k →∞,
and similarly u
γ−1+p(x)
k
→ u
γ−1+p(x)
monotonically when γ −1+p(x) < 0. If γ −1+p(x) ≥
0, we have
u
γ−1+p(x)
k



∇
η


p(x)
≤ C

1+u
γ−1+p(x)



∇
η


p(x)
≤ C

2+u
p(x)



∇
η


p(x)
.

(4.12)
since γ is negative. Now we can let k
→∞in the above inequality, obtaining the claim by
the monotone converge theorem and the dominated converge theorem.

In the following two theorems, q is an exponent such that 1 <q<n/(n −1), s>p
+
B
4R
−
p
−
B
4R
,andq
0
> 0 is an exponent for which the weak Harnack inequality holds for the
function under consideration.
Theorem 4.4. Let u be a nonnegative supersolution in B
4R
, B
2r
⊂ B
4R
, γ<γ
0
< 0, η ∈
C
∞
0

(B
4R
) such that 0 ≤η ≤ 1 and |∇η|≤C/r. Then

B
r
∩{u≤λ}


∇
u


p(x)
η
p
+
B
4R
dx ≤ Cλ
1−γ
r
n−p
−
B
2r

essinf
B
r

u + r

(γ−1+p
−
B
2r
)
, (4.13)
16 Boundary Value Problems
where γ is chosen so that q
0
≥ q(γ −1+p
−
B
2r
) > 0 and the constant C depends on n, p, γ
0
,
and the L
q

s
(B
4R
)-norm of u.
Proof. We h ave u/λ
≤ 1wheneveru ≤ λ. Using this fact, Lemmas 3.3, 3.4,and4.3,the
H
¨
older inequalit y, and the weak Harnack inequality we obtain


B
r
∩{u≤λ}
|∇u|
p(x)
η
p
+
B
4R
dx
≤

B
r
∩{u≤λ}

u
λ

γ−1


∇
u


p(x)
η

p
+
B
4R
dx
≤ Cλ
1−γ

B
2r
u
γ+p(x)−1


∇
η


p(x)
dx
≤ Cλ
1−γ
r
−p
−
B
2r

B
2r

u
γ−1+p(x)
dx
≤ Cλ
1−γ
r
n−p
−
B
2r

−

B
2r
u
q

(p(x)−p
−
B
2r
)
dx

1/q


−


B
2r
u
q(γ−1+p
−
B
2r
)
dx

1/q
≤ Cλ
1−γ
r
n−p
−
B
2r

1+u
q

(p
+
B
4R
−p
+
B
4R

)
L
q

s
(B
4R
)

1/q


−

B
2r
u
q(γ−1+p
−
B
2r
)
dx

1/q
≤ Cλ
1−γ
r
n−p
−

B
2r

essinf
B
r
u + r

(γ−1+p
−
B
2r
)
.
(4.14)

The Sobolev p(·)-capacity of a set E ⊂ R
n
is defined as
C
p(·)
(E) = inf

R
n



u(x)



p(x)
+


∇
u(x)


p(x)

dx, (4.15)
where the infimum is taken over the set of admissible functions
S
p(·)
(E) =

u ∈ W
1,p(·)
(R
n
):u ≥ 1inanopensetcontainingE

. (4.16)
This definition gives a Choquet capacity; for this and other properties of C
p(·)
,see[10].
The following theorem is our main result.
Theorem 4.5. Let u be a nonnegative function such that
(1) u is lower semicontinuous,

(2) min
{u,λ} is a supersolution for each λ>0,and
(3) u
∈ L
t
loc
(Ω) for some t>0.
Denote
E
λ
=

x ∈ B

x
0
,r

: u(x) >λ

, (4.17)
PetteriHarjulehtoetal. 17
where B(x
0
,r) is a ball with B
4r
= B(x
0
,4r)Ω. Then
C

p(·)

E
λ

≤
Cr
n−p
−
B
2r
λ
−q
0
/q

inf
B
r
u + r

q
0
/q
, (4.18)
where the constant C depends on p, n,andtheL
t
(B
4r
)-norm of u.

Remark 4.6. (1) Observe that all supersolutions satisfy the assumptions of the previous
theorem.
(2) For constant p(
·), the class of functions which satisfy (1)–(3) is called p-
superharmonic functions. This is a strictly bigger class of functions than supersolutions.
Indeed, the nonlinear counterpart of a fundamental solution is the prime example of such
a function.
Proof. Denote u
λ
= min{u,λ} and choose ϕ ∈ C
∞
0
(B(x
0
,2r)) such that 0 ≤ ϕ ≤ 1, ϕ = 1
in B(x
0
,r), and |∇ϕ|≤C/r.Forsufficiently small radii r,wecanchoose
q
0
=
C
C + u
p
+
B
4R
−p
−
B

4R
L
t
(B
4R
)
(4.19)
since we can take q

s = t with a suitable choice of s>p
+
B
4r
− p
−
B
4r
. Then the weak Harnack
inequality holds for u
λ
with an exponent and a constant independent of λ.Further,we
choose the parameter γ in Theorem 4.4 so that q
0
/q =γ −1+p
−
B
2r
. This is always possible,
since we can take a smaller q
0

if necessary.
Since u is lower semicontinuous, the set E
λ
is open. Further, u
λ
ϕ/λ = 1inE
λ
,sowecan
test the capacity of E
λ
with u
λ
ϕ/λ. This gives
C
p(·)
(E
λ
) ≤

B(x
0
,2r)





u
λ
ϕ

λ




p(x)
+




∇

u
λ
ϕ

λ




p(x)

dx
≤ λ
−p
−
B
2r


B(x
0
,2r)



u
λ
ϕ


p(x)
+


∇

u
λ
ϕ



p(x)

dx.
(4.20)
For the first term in the integral above we have by the H
¨

older inequality, Lemma 3.4,and
the weak Harnack inequality that

B(x
0
,2r)


u
λ


p(x)
dx = Cr
n
−

B(x
0
,2r)


u
λ


p(x)−p
−
B
2r



u
λ


p
−
B
2r
dx
≤ Cr
n

−

B(x
0
,2r)


u
λ


q

(p(x)−p
−
B

2r
)

1/q


−

B(x
0
,2r)


u
λ


qp
−
B
2r
dx

1/q
≤ Cr
n

1+u
q


(p
+
B
4r
−p
−
B
4r
)
L
q

s
(B
4r
)

−

B(x
0
,2r)


u
λ


qp
−

B
2r
dx

1/q
≤ Cr
n
λ
p
−
B
2r
−q
0
/q

−

B(x
0
,2r)


u
λ


q
0
dx


1/q
≤ Cr
n
λ
p
−
B
2r
−q
0
/q

inf
B(x
0
,r)
u + r

q
0
/q
.
(4.21)
18 Boundary Value Problems
For the second term, we get by the product rule


∇


u
λ
ϕ



p(x)
≤ C



ϕ∇u
λ


p(x)
+


u
λ
∇ϕ


p(x)

. (4.22)
Using Lemma 3.3 and estimating the average of
|u
λ

|
p(x)
as above we have

B(x
0
,2r)


u
λ
∇ϕ


p(x)
dx ≤ Cr
n−p
−
B
2r

−

B(x
0
,2r)


u
λ



p(x)
dx

≤
Cr
n−p
−
B
2r
λ
p
−
B
2r
−q
0
/q

inf
B(x
0
,r)
u + r

q
0
/q
.

(4.23)
We estimate the remaining term by using Theorem 4.4. To this end, choose a function
η
∈ C
∞
0
(B
3r
)suchthat0≤η ≤ 1, η = 1inB(x
0
,2r), and |∇η|≤C/r.Then

B(x
0
,2r)


ϕ∇u
λ


p(x)
dx ≤

B(x
0
,2r)


∇

u
λ


p(x)
dx
=

B(x
0
,2r)


∇
u
λ


p(x)
η
p
+
B
4r
dx
≤ Cλ
1−γ
r
n−p
−

B
2r

inf
B
r
u + r

q(γ−1+p
−
B
2r
)
.
(4.24)
Combining the above estimates we have
C
p(·)
(E
λ
) ≤ Cr
n−p
−
B
2r

λ
−q
0
/q

+ λ
1−γ−p
−
B
2r


inf
B
r
u + r

q
0
/q
. (4.25)
Now our choice of γ gives the claim.

The previous result implies that the singularity set of a supersolution is of zero capac-
ity.
Corollary 4.7. For functions u satisfying the assumptions of Theorem 4.5,
C
p(·)
({x ∈ Ω : u(x) =∞}) =0. (4.26)
Proof. Fix a ball B(x
0
,r)asinTheorem 4.5 and let
E
i
=


x ∈ B

x
0
,r

: u(x) >i

, i = 1,2, ,
E
=

x ∈ B

x
0
,r

: u(x) =∞

.
(4.27)
PetteriHarjulehtoetal. 19
Since E
=∩
i
E
i
and E

1
⊃ E
2
⊃···, we get by the monotonicity of the capacit y and
Theorem 4.5 that
C
p(·)
(E) ≤ lim
i→∞
C
p(·)

E
i

=
0. (4.28)
Since Ω can be covered by a countable number of balls for which (4.28)holds,thesub-
additivity of the capacity implies the claim.

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Petteri Harjulehto: Department of Mathematics and Statistics, University of Helsinki, P.O. Box 68,
Helsinki, Finland
Email address: petteri.harjulehto@helsinki.fi
Juha Kinnunen: Department of Mathematical Sciences, University of Oulu, P.O. Box 3000,
Oulu, Finland
Email address: juha.kinnunen@oulu.fi
Teemu Lukkari: Institute of Mathematics, Helsinki University of Technology, P.O. Box 1100,
Espoo, Finland
Email address: teemu.lukkari@tkk.fi